A student standardized a NaOH solution, following the Procedure described in thi

Question

A student standardized a NaOH solution, following the Procedure described in this module. The student weighted two samples of KHP and transferred them into two 125-ml. She dissolved each sample, using distilled water. After adding phenolphthalein indicator solution to each KHP solution, she titrated the solutions with the NaOH solution being standardized. The student's data are given in Table 1. Complete the following calculations and enter the answers in Table 1. Express the volume of NaOH solution used for each titration in liters. Calculate the number of moles of KHP titrated in each determination.

Explanation / Answer

Solution :-

Following are the steps for the calculation of the each part so that you can plug in the values in the table

Molar mass of KHP = 204.22 g/mol

Lets calculate the moles of KHP in each part

Moles = mass in gram / molar mass

Moles of KHP trial 1 = 3.009 g /204.22 g per mol = 0.01473 mol KHP

Moles of KHP trial 2 = 3.246 g / 204.22 g per mol = 0.01589 mol KHP

Volume of NaOH required in L

Trial 1 = 42.83 ml * 1 L / 1000 ml = 0.04283 L

Trial 2 = 45.93 ml * 1 L / 1000 ml = 0.04593 L

Moles of KHP titrated trial 1 = 0.01473 mol KHP

Moles of KHP titrated trial 2 = 0.01589 mol KHP

Moles of NaOH required for titration are same as moles of KHP because mole ratio is 1: 1

So moles of NaOH required trial 1 = 0.01473 mol

moles of NaOH required trial 2 = 0.01589 mol

molarity of NaOH= moles / volume in L

trial 1 molarity of NaOH = 0.01473 mol /0.04283 L = 0.344 M

molarity of NaOH trial 2 = 0.01589 mol / 0.04593 L= 0.346 M

Mean molarity of NaOH = (0.344 M+ 0.346 M) /2 = 0.345 M

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